In particular $F\_\{\backslash min\}(\backslash alpha)$ determines $\backslash mathcal\{F\}[\backslash alpha]$ , and furthermore $\backslash mathcal\{F\}[\backslash alpha]$ determines $F\_\{\backslash min\}(\backslash alpha)$ up to conjugacy except in the case that $\backslash mathcal\{F\}[\backslash alpha]=\backslash mathcal\{A\}$ : in that case, $F\_\{\backslash min\}(\backslash alpha)$ can be either trivial or atomic. Despite this ambiguity we will refer to either the free factor $F\_\{\backslash min\}(\backslash alpha)$ or the free factor system $\backslash mathcal\{F\}[\backslash alpha]$ as the filling support of $\backslash alpha$ ; in context the exact referent should be clear.

We sometimes use the notation $\backslash mathcal\{F\}[\backslash alpha;T]$ to highlight the dependence on the free splitting $T$ that contains $\backslash alpha$ . Also, we note that $\backslash mathcal\{F\}[\backslash alpha]$ is determined by the connected subgraph $\backslash beta\_\{\backslash mathrm\{Id\}\}^\{\backslash circ\}(\backslash alpha)$ , in that it is the unique smallest free factor system that "carries" the subgroup $\backslash mathrm\{Stab\}(\backslash beta\_\{\backslash mathrm\{Id\}\}^\{\backslash circ\}(\backslash alpha))$ . For this reason we sometimes denote $\backslash mathcal\{F\}[\backslash alpha]=\backslash mathcal\{F\}[\backslash beta^\{\backslash circ\}\_\{\backslash mathrm\{Id\}\}(\backslash alpha;T)]$ and $\backslash mathrm\{KR\}(\backslash alpha)=\backslash mathrm\{KR\}(\backslash beta^\{\backslash circ\}\_\{\backslash mathrm\{Id\}\}(\backslash alpha;T))$ .

Remark. In the statement of Definition 2.8, Case (1) holds if and only if $\backslash beta\_\{\backslash mathrm\{Id\}\}^\{\backslash circ\}(\backslash alpha)=\backslash alpha$ if and only if for all $g\backslash neq\; h\backslash in\backslash Gamma$ the translated paths $g\backslash cdot\backslash alpha$ and $h\backslash cdot\backslash alpha$ do not overlap; in this case the protocomponents of $\backslash beta^\{\backslash circ\}(\backslash alpha)$ are precisely the translates of $\backslash alpha$ . Also, Case (2) holds if and only if there exists a vertex $V\backslash in\backslash beta\_\{\backslash mathrm\{Id\}\}^\{\backslash circ\}(\backslash alpha)$ such that $\backslash mathrm\{Stab\}(\backslash beta\_\{\backslash mathrm\{Id\}\}^\{\backslash circ\}(\backslash alpha))=\backslash mathrm\{Stab\}(V)$ and $[\backslash mathrm\{Stab\}(V)]\backslash in\backslash mathcal\{A\}$ ; it follows in this case that $V\backslash in\backslash alpha$ , and that $\backslash beta\_\{\backslash mathrm\{Id\}\}^\{\backslash circ\}(\backslash alpha)=\backslash bigcup\_\{g\backslash in\backslash mathrm\{Stab\}(V)\}g\backslash cdot\backslash alpha$ .

A continued example.

In the example we gave preceding the statement of Theorem 2.4, the protocomponents of $\backslash beta\_\{\backslash mathrm\{Id\}\}^\{\backslash circ\}(\backslash alpha)$ are the axes in $T$ of the infinite cyclic free factor $\backslash langle\; a\backslash rangle$ of $\backslash Gamma=\backslash langle\; a,b\backslash rangle$ , and so the filling support is $\backslash mathcal\{F\}[\backslash alpha;T]=\backslash \{[\backslash langle\; a\backslash rangle]\backslash \}$ which is a nonfull free factor system, despite the fact that the path $\backslash alpha\backslash subset\; T$ does indeed have an interior crossing of a translate of every natural edge orbit of $T$ : the tree $T$ has only one natural edge orbit, and $\backslash alpha$ crosses three different natural edges of that orbit, the middle one of those three being an interior crossing.

A new example.

This example shows that the subgroup $\backslash mathrm\{Stab\}(\backslash beta\_\{\backslash mathrm\{Id\}\}^\{\backslash circ\}(\backslash alpha))$ need not be a free factor of $\backslash Gamma$ rel $\backslash mathcal\{A\}$ , in contrast to the fact that the stabilizer of every component of the $\backslash Gamma$ -invariant subforest $\backslash beta(\backslash alpha)\backslash subset\; T$ is a free factor of $\backslash Gamma$ rel $\backslash mathcal\{A\}$ . Consider the rank 4 free group $\backslash Gamma=\backslash langle\; a,b,c,d\backslash rangle$ . Let $G$ be a marked graph with two vertices $p,q$ , with the $\backslash langle\; a,b\backslash rangle$ rose attached to $p$ , the $\backslash langle\; c,d\backslash rangle$ rose attached to $q$ , and an edge $e$ from $p$ to $q$ ; we identify $\backslash pi\_1(G,p)\backslash approx\backslash langle\; a,b\backslash rangle*\backslash langle\; c,d\backslash rangle\backslash approx\backslash langle\; a,b,c,d\backslash rangle$ by using $e$ as a maximal subtree of $G$ . Let $U=\backslash widetilde\{G\}$ be the universal covering with edge labels lifted from $G$ . Choosing a lift $\backslash tilde\{p\}\backslash in\; U$ of $p$ determines the deck transformation action $\backslash Gamma\backslash curvearrowright\; U$ . Let $\backslash tilde\{\backslash alpha\}\backslash subset\; U$ be the path with initial vertex $\backslash tilde\{p\}$ that is labelled by the word $ecd\backslash bar\{c\}\backslash bar\{d\}\backslash bar\{e\}ab\backslash bar\{a\}\backslash bar\{b\}e$ . Note that $\backslash tilde\{\backslash alpha\}$ is contained in the $U$ -axis of the infinite cyclic subgroup

Also, $\backslash tilde\{\backslash alpha\}$ consists of one entire fundamental domain for the action of $C$ on its $U$ -axis, followed by the first $e$ edge of the next fundamental domain. Let $T$ be the free splitting obtained from $U$ by collapsing all edges labelled $a,b,c,d$ , so $T$ has a single natural edge orbit, represented by the $T$ -image of any $e$ -edge of $U$ . Let $\backslash alpha\backslash subset\; T$ be the image of $\backslash tilde\{\backslash alpha\}$ . Note that $\backslash alpha$ crosses the unique natural edge orbit three times, the middle crossing being an interior crossing. Again $\backslash alpha$ is contained in the axis of $C$ in $T$ , and $\backslash alpha$ consists of one entire fundamental domain of that axis followed by the first edge of the next fundamental domain. Note that no two distinct translates of the axis of $C$ have a common edge in $T$ . It follows that the protocomponent $\backslash beta\_\{\backslash mathrm\{Id\}\}^\{\backslash circ\}(\backslash alpha)$ equals the axis of $C$ in $T$ , and that $C=\backslash mathrm\{Stab\}(\backslash beta\_\{\backslash mathrm\{Id\}\}^\{\backslash circ\}(\backslash alpha))$ . But $C$ has trivial image under abelianization of $\backslash langle\; a,b,c,d\backslash rangle$ and hence $C$ is not a free factor of $\backslash langle\; a,b,c,d\backslash rangle$ . In fact we have the equation $\backslash mathcal\{F\}[\backslash alpha;T]=\backslash \{[\backslash langle\; a,b,c,d\backslash rangle]\backslash \}$ , in other words $C$ fills the group $\backslash langle\; a,b,c,d\backslash rangle$ , and hence $\backslash alpha$ fills $T$ by Proposition 2.9.

The fact that $C$ fills $\backslash langle\; a,b,c,d\backslash rangle$ can be proved using a beautiful transversality argument that we learned from a paper of Stallings [Sta00]. Here are some details of this argument. Consider $S$

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